Question

2. A metal block of area 0-20 m² is connected to a
0-020 kg mass via a string that passes over an ideal
pulley (considered massless and frictionless), as
in Fig. 7(c) 20. A liquid with a film thickness of
0-40 mm is placed between the block and the table.
When released the block moves to the right with a constant speed of 0.08 m/s .
Find the coefficient of viscosity
of the liquid.​

Answer 1

Answer:

3.5×10

−3

Pass

Explanation:

Here, m=0.01kg,

l=0.3mm=0.3×10

−3

m,

g=10ms

−2

,

v=0.085ms

−1

,

A=0.1m

2

.

The metal block moves to the right due to tension T of the string which Is equal to the weight of the mass suspended at the end of the string.

Thus,

Shear force, F=T=mg

=0.01kg×10ms

−2

=0.1N

Shear stress on the fluid =

A

F

=

0.1m

2

0.1N

Strain rate =

l

v

=

0.3×10

−3

m

0.085ms

−1

Coefficient of viscosity, η=

Strainrate

Shearstress

=(

0.1m

2

0.1N

)×

0.085ms

−1

0.3×10

−3

m

=3.5×10

−3

Pas